Physics

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A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 35.0 m above sea level, directed at an angle theta = 45.9° above the horizontal, and with a speed v = 32.3 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.

  • Physics -

    Vo = 32.3m/ @ 45.9o.
    Xo = 32.3*cos45.9 = 22.5 m/s.
    Yo = 32.3*sin45.9 = 23.20 m/s.

    Y = Yo + g*t.
    Tr = (Y-Yo)/g = (0-23.2)/-9.8 = 2.37 s.
    = Rise time.

    Tf1 = Tr = 2.37 s. = Fall time from hmax
    to 35 m above sea level.

    h = Yo*t + 0.5g*t^2 = 35 m.
    23.2*t + 4.9t^2 = 35
    4.9t^2 + 23.2t - 35 = 0.
    Use Quadratic Formula and get:
    Tf2 = 1.20 s. = Fall time from 35 m above sea level to sea level.

    T = Tr + Tfi + Tf2.
    T = 2.37 + 2.37 + 1.20 = 5.94 s. = Time
    in air.

    D = Xo * T = 22.5m/s * 5.94s = 133.7 m.

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